ZOJ 1489 2^x mod n = 1

这题想了很久才想出一个做法

在纸上乱写着写着，就出来了

自己也不知道为什么对

2^x mod n = 1

 1 #include<iostream>
 2 #include <cmath>
 3  using namespace std;
 4  int isfindtwo(int);
 5 
 6  int main()
 7 {
 8     int n;
 9     int count = 1;
10     int hold;
11     int a;
12     //freopen("test.txt","r", stdin);
13     while (cin >> n)
14     {
15         if (n % 2 == 0 || n == 1)
16             cout << "2^? mod " << n << " = 1" << endl;
17     
18         else
19         {
20             if (isfindtwo(n+1))
21                 cout << "2^" << isfindtwo(n+1) << " mod " << n << " = 1" << endl;
22             else
23             {
24                 hold = (n+1)/2;
25                 a = hold + n;
26                 while (!isfindtwo(a))
27                 {
28                     if (a%2 == 0)
29                     {    
30                         a /= 2;
31                         count++;
32                     }
33                     else
34                         a += n;
35                 }
36                 cout << "2^" << count+isfindtwo(a) << " mod " << n << " = 1" << endl;
37             }
38         }
39         count = 1;
40     }
41     return 0;
42 }
43 //查找n是不是2的指数的形式
44 int isfindtwo(int n)
45 {
46     for (int i = 1; ; i++)
47     {
48         if ((int)pow((double)2,(int)i) == n)
49         {
50             return i;
51         }
52         else
53         {
54             if ((int)pow((double)2,(int)i) > n)
55             {
56                 return 0;
57             }
58         }
59     }
60 }

 

之后在网上找找别人是如何做的，受到启发，重写了一个

主要用到一个公式 (m*n)%d =( m%d * n%d ) %d

 

2^x mod n = 1

#include<iostream>
#include <cmath>
using namespace std;
int isfindtwo(int);

int main()
{
int n;
int count = 0;
int hold;
int a = 1;
//freopen("test.txt","r", stdin);
    while (cin >> n)
    {
if (n % 2 == 0 || n == 1)
            cout << "2^? mod " << n << " = 1" << endl;

else
        {
while (1)
            {
                a <<= 1;
                count++;
if (a > n)
                    a %= n;

if (a % n == 1)
break;
            }                
            cout << "2^" << count << " mod " << n << " = 1" << endl;

        }
        count = 0;
    }
return 0;
}